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y=$x^4$+4

x=any random 5digit natural no.. Find probability that y is divisible by 5?

options:

a)$1$/$5$ b)$4$/$5$ c)$8$/$9$

MyApproach

Total possible outcomes(a)= $8$ . $9$ . $9$ .$9$ . $9$=

Total favourable outcomes(b)=$8$ . $9$ . $9$ .$9$ . $2$=

P(getting $5$) =a/b=$9$/$2$.

MyApproach2

From $1$ to $9$ I can put x values as $1$,$2$,$3$,$6$,$7$,$9$ that on adding 4 will be divisible by 5.

So,From $1$ to $10$ I have $6$ values of x .When added by $4$ these values are divisible by $5$.

So,$10$ can have $6$ values.

$1$ can have $6/10$ values.

$99,998$ can have=$99,998$ . $6/10$ values.

I got this $99,998$ as 1+n-1=$99999$=n=99,998

But I am not getting any correct Ans.

Can anyone guide me how to solve the problem?

Jack
  • 752

4 Answers4

3

Consider the following cases:

  • $x\equiv0\pmod5 \implies x^4+4\equiv0^4+4\equiv 4\equiv4\pmod5$
  • $x\equiv1\pmod5 \implies x^4+4\equiv1^4+4\equiv 5\equiv0\pmod5$
  • $x\equiv2\pmod5 \implies x^4+4\equiv2^4+4\equiv 20\equiv0\pmod5$
  • $x\equiv3\pmod5 \implies x^4+4\equiv3^4+4\equiv 85\equiv0\pmod5$
  • $x\equiv4\pmod5 \implies x^4+4\equiv4^4+4\equiv260\equiv0\pmod5$

In $4$ out of $5$ cases, $x^4+4\equiv0\pmod5$, hence the probability is $\dfrac45$.

barak manos
  • 43,109
  • Please note that it works "accurately" only for $x\in[5m\dots5n-1]$. Since you specified $x\in[10000\dots100000-1]$, it works "accurately". – barak manos Nov 08 '15 at 13:13
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Hint:

Find $x^4+4$ for some small values like $0,1,2,3,4$.

If $x=5n+k$ where $k\in\{0,1,2,3,4\}$ then expanding $x^4+4$ leads to:

$$x^4+4=5m+k^4+4$$

That shows that: $$5|x^4+4\iff 5|k^4+4$$

So checking for $k\in\{0,1,2,3,4\}$ is enough to get a picture of the whole.

drhab
  • 151,093
  • I have edited the code.I already did for x values of 1,2,3,6,7,9 – Jack Nov 08 '15 at 12:43
  • Can you find a pattern if it comes to divisibility by $5$? – drhab Nov 08 '15 at 12:45
  • Ya for every 10 values I get 6 values that are divisible by 5.I have already said that please read the solution sir. – Jack Nov 08 '15 at 12:46
  • $x^4+4$ is not divisible by $5$ if and only if $x$ is divisible by $5$. So for $10$ consecutive values $8$ are divisible by $5$. – drhab Nov 08 '15 at 12:49
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Here's a hint: If $\text{gcd}(a,5)=1$, then $a^4\equiv 1\pmod{5}$. This means that $5$ divides $a^4-1$, or that $5$ divides $a^4+4$.

Edit

There is this result called Fermat's little theorem that says that if $5$ does not divide $a$, then $5$ will divide $a^4-1$. This means for instance that when $a$ is among the numbers $\{1,2,3,4\}$, then $5$ divides $a^4-1$.

It is also true that if $5$ does divide $a$, then $5$ will not divide $a^4-1$.

If $5$ divides something, then you can add $5$ to that something and still have that $5$ divides it. In our case, if $5$ divides $a^4-1$, then $5$ will also divide $a^4-1+5=a^4+4$.

In total, it looks like only the numbers $a$ that $5$ does not divide have the property that $5$ divides $a^4+4$, so you need to think about the fraction of $5$-digit numbers that $5$ does not divide.

Mankind
  • 13,170
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Hint: $x^4+4$ can be written as $x^4-1+5$. If $x^4-1+5$ is divisible by 5 then $x^4-1$ is divisible by 5. Using th third binomial formula $(x^2-1)\cdot (x^2+1)$. Insert the digits $0, ... ,9$. If at least one of the two factors is divisible by 5, then $x^4+4$ is divisible by 5.

callculus42
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